MAT Mathematical Modelling [ Pemodelan Matematik]
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1 UNIVERSITI SAINS MALAYSIA First Semester Examination 2016/2017 Academic Session December 2016/January 2017 MAT Mathematical Modelling [ Pemodelan Matematik] Duration : 3 hours [Masa : 3 jam] Please check that this examination paper consists o TEN pages o printed material beore you begin the examination. [Sila pastikan bahawa kertas peperiksaan ini mengandungi SEPULUH muka surat yang bercetak sebelum anda memulakan peperiksaan ini.] Instructions: Answer all our [4] questions. [Arahan: Jawab semua empat [4] soalan.] In the event o any discrepancies, the English version shall be used. [Sekiranya terdapat sebarang percanggahan pada soalan peperiksaan, versi Bahasa Inggeris hendaklah diguna pakai]....2/
2 a Explain about the no-slip condition in a luid low. b Discuss briely on the classiications o luid low. [14 marks] 1. a Terangkan tentang syarat tak gelincir dalam aliran bendalir. b Bincangkan secara ringkas mengenai pengelasan-pengelasan aliran bendalir. [14 markah] 2. Assume that the momentum and energy equations with constant surace heat rate or axisymmetric low in a circular tube o radius a in x-direction is given by the ollowing equation, respectively: 1 u r 2 µ u r 2 α T = r 2 dp dx, 1 = r 2 dt m dx, 2 where x and r are the axial and radial coordinates o circular tube, respectively, and u and T are the ully developed velocity and temperature proiles, respectively. The pressure P and the mass-averaged luid temperature T m are independent o r, and the dynamic viscosity µ and the thermal diusivity α o the luid are constants. a Derive a ully developed temperature proile T by using the momentum equation 1 and the energy equation 2 subject to the ollowing boundary conditions u = 1 xr, T = 2 u 3 at r = 0, u = 0, T = T w at r = a. [ T w is the luid temperature at the surace o circular tube. ]...3/-
3 - 3 - b What is the velocity gradient near the surace o the tube? c What is the temperature proile at the centerline o the tube? [26 marks] 2. Anggapkan bahawa persamaan-persamaan momentum dan tenaga dengan kadar haba permukaan malar bagi aliran simetri sepaksi dalam tiub bulat berjejari a dalam arah x masing-masing diberikan oleh persamaan-persamaan berikut: 1 u r 2 µ u r 2 α T = r 2 dp dx, 1 = r 2 dt m dx, 2 dengan x dan r masing-masing ialah koordinat paksian dan jejarian tiub bulat, dan u dan T masing-masing ialah proil halaju dan suhu terbangun penuh. Tekanan P dan suhu jisim-purata bendalir T m tidak bersandar pada r, dan kelikatan dinamik µ dan resapan terma α adalah malar. a Terbitkan proil suhu terbangun penuh T dengan menggunakan persamaan momentum 1 dan persamaan tenaga 2 tertakluk kepada syarat-syarat sempadan berikut u = 1 xr, T = 2 u 3 pada r = 0, u = 0, T = T w pada r = a. [ T w ialah suhu bendalir pada permukaan tiub bulat. ] b Apakah kecerunan halaju berdekatan permukaan tiub? c Apakah proil suhu pada garis tengah tiub? [26 markah]...4/-
4 Consider a steady low along a semi-ininite two-dimensional surace with a ree stream temperature T and a ree stream velocity u where x is measured along the surace and y normal to the surace. An ininitesimal stationary control volume o unit depth within the boundary layer with various rates o energy transer across the control surace o the luid mixture is shown in Figure 1. di, y+ W conv, y+ shear, y+ q y + v E conv, x u conv, x+ δ x δ x E conv, y q y W shear, y E di, y Figure 1: Control volume and energy transer terms or development o the steady-low energy dierential equation o the boundary layer Table 1: Basic rates o energy transer Ė conv,x = M x δy i u2 convection rate: assuming u 2 v 2, Ė di,y = m j γ j j i j δ x diusion rate without Soret eect, T q y = k δ x conduction heat transer without Duour eect, Ẇ shear,y = τ yx uδx shear orce velocity a What is the principle o conservation o energy? b The two-dimensional boundary layer continuity and momentum equations or this case are given, respectively, as ollows: M x x + M y = 0, 3 u M x x + M u y = dp dx + µ u /-
5 - 5 - Based on Figure 1, Table 1 and applying the principle rom 3a, the two dimensional boundary layer approximations, equations 3 and 4, show that the energy equation o the boundary layer is ρu i i + ρv x = k T + m j γ j j i j + µ u 2 + u dp dx. 5 Then, state the assumption that can be made so that the mass diusion term in equation 5 can be neglected. [ Note: u,v = velocity components along x,y axes; M x,m y = ρu,ρv where M represent the mass lux and ρ = the density o the luid; i = the mixture enthalpy; i j = the partial enthalpy o component j; γ j = mass diusion coeicient or component j in the mixture; m j = mass concentration o component j in the mixture; k = the thermal conductivity; T = temperature o the luid mixture; τ = shear stress ] [25 marks] 3. Pertimbangkan suatu aliran mantap di sepanjang permukaan dua dimensi separuh tak terhingga dengan suhu strim bebas T dan halaju strim bebas u dengan x diukur di sepanjang permukaan dan y serenjang terhadap permukaan. Satu unit kedalaman unsur isipadu kawalan pegun dalam lapisan sempadan dengan beberapa kadar pemindahan tenaga merentasi permukaan kawalan bendalir campuran ditunjukkan dalam Rajah 1. conv, y+ di, y+ q y + W shear, y+ v E conv, x u conv, x+ δ x δ x E conv, y q y W shear, y E di, y Rajah 1: Isipadu kawalan dan sebutan-sebutan pemindahan haba untuk menerbitkan persamaan pembezaan tenaga bagi aliran lapisan sempadan yang mantap...6/-
6 - 6 - Jadual 1: Kadar-kadar asas pemindahan tenaga Ė conv,x = M x δy i u2 kadar olakan: andaian u 2 v 2, Ė di,y = m j γ j j i j δx kadar resapan tanpa kesan Soret, T q y = k δx pemindahan haba konduksi tanpa kesan Duour, Ẇ shear,y = τ yx uδx daya ricih halaju a Apakah prinsip keabadian tenaga? b Persamaan-persamaan keselanjaran dan momentum bagi lapisan sempadan dua dimensi untuk kes ini masing-masing diberikan seperti berikut: M x x + M y = 0, 3 u M x x + M u y = dp dx + µ u. 4 Berdasarkan Rajah 1, Jadual 1 dan menggunakan prinsip daripada 3a, penghampiran lapisan sempadan dua dimensi, persamaan-persamaan 3 dan 4, tunjukkan bahawa persamaan tenaga lapisan sempadan ialah ρu i i + ρv x = k T + m j γ j j i j + µ u 2 + u dp dx. 5 Seterusnya, nyatakan andaian yang boleh dibuat supaya sebutan resapan jisim dalam persamaan 5 boleh diabaikan. [ Nota: u,v = komponen-komponen halaju sepanjang paksi x,y; M x,m y = ρu,ρv dengan M mewakili luks jisim dan ρ = ketumpatan bendalir; i = entalpi campuran; i j = entalpi separa komponen j; γ j = pekali resapan jisim komponen j dalam campuran; m j = kepekatan jisim komponen j dalam campuran; k = kekonduksian terma; T = suhu bendalir campuran; τ = tegasan ricih ] [25 markah]...7/-
7 Consider a steady mixed convection boundary layer low past a horizontal circular cylinder o radius a with constant surace heat lux q w and is placed in a constant ree stream temperature T. It is assumed that the constant ree stream velocity is 1/2U so that the velocity outside the boundary layer is ū e = U sin x/a. It is also assumed that the ree stream velocity is directed vertically upward with q w > 0 or assisting low and q w < 0 or opposing lows. The cylinder is considered to be long enough so that the end eects can be neglected and accordingly the low ield can be assumed two-dimensional. Under these assumptions along with the Boussinesq and boundary layer approximations, the dimensional basic equations are ū x + v ȳ ū ū x + v ū ȳ ū T x + v T ȳ = 0, 6 = ū e dū e d x + υ 2 ū ȳ 2 + gβ T T sin x/a, 7 = υ 2 T Pr ȳ 2, 8 subject to the boundary conditions T ū = v = 0, ȳ = q w at ȳ = 0, 9 k ū ū e, T T as ȳ, where x and ȳ are the Cartesian coordinates measured along the surace o the cylinder starting rom the lower or upper stagnation points and normal to it, respectively, ū and v are velocity components along x and ȳ axes, respectively, T is the temperature o the luid, υ is the kinematic viscosity o the luid, g is the gravitational acceleration, β is the thermal expansion coeicient o the luid, Pr is the Prandtl number and k is the thermal conductivity o the luid. The non-dimensional variables are x = x/a, y = Re 1/2 ȳ/a, u = ū/u, v = Re 1/2 v/u, u e x = ū e /U, T = Re 1/2 T T q w a/k, λ = Gr Re 5/2, where Re = U a/υ is the Reynolds number, Gr = gβq w a 4 /kυ 2 is the Grasho number and λ is the mixed convection parameter. It is noticed that λ > 0 or assisting low and λ < 0 is or opposing lows. Note: ψ is a stream unction which is deined as u = ψ and v = ψ x. a By using non-similarity variables o the ollowing orm ψ = x x,y, T = θx,y,...8/-
8 - 8 - show that the boundary layer equations 6 8 can be reduced to the ollowing system o dierential equations sinxcosx + λ sinx x x θ 2 = x x 2 x 2, θ Pr 2 + θ = x θ x θ, 11 x with the boundary conditions 9 becoming = = 0, θ = 1 at y = 0, 12 sinx, g 0 as y. x 2 b Table 2 displays the values o shear stress 2 and surace temperature θ y=0 or the system o equations 10 and 11 subject to boundary y=0 conditions 12 when x = 0 and Pr = 7 with various values o λ. Interpret the obtained results in Table 2. Table 2: Values o shear stress and surace temperature with various values o λ λ 2 2 y=0 θ y= [35 marks] 4. Pertimbangkan suatu aliran lapisan sempadan olakan campuran yang mantap terhadap silinder bulat mendatar berjejari a dengan luks haba permukaan malar q w dan diletakkan dalam suhu strim bebas malar T. Andaikan bahawa halaju strim bebas malar ialah 1/2U supaya halaju luar lapisan sempadan ialah ū e = U sin x/a. Andaikan juga bahawa halaju strim bebas adalah dalam arah tegak ke atas dengan q w > 0 untuk aliran membantu dan q w < 0 untuk aliran menentang. Silinder dianggap panjang supaya kesan-kesan hujung boleh diabaikan dan dengan demikian, medan...9/-
9 - 9 - aliran boleh diandaikan dalam dua dimensi. Berdasarkan andaian-andaian tersebut beserta penghampiran-penghampiran Boussinesq dan lapisan sempadan, persamaanpersamaan asas berdimensi ialah ū x + v ȳ = 0, 6 ū ū x + v ū ȳ ū T x + v T ȳ = ū e dū e d x + υ 2 ū ȳ 2 + gβ T T sin x/a, 7 = υ 2 T Pr ȳ 2, 8 tertakluk kepada syarat-syarat sempadan T ū = v = 0, ȳ = q w pada ȳ = 0, 9 k ū ū e, T T apabila ȳ, dengan x dan ȳ masing-masing ialah koordinat-koordinat Cartesan yang diukur di sepanjang permukaan silinder bermula dari titik genangan bawah atau atas dan berserenjang dengannya, ū dan v masing-masing ialah komponen-komponen halaju pada paksi x dan ȳ, T ialah suhu bendalir, υ ialah kelikatan kinematik bendalir, g ialah pecutan graviti, β ialah pekali pengembangan terma bendalir, Pr ialah nombor Prandtl dan k ialah kekonduksian terma bendalir. Pemboleh-pemboleh ubah tak berdimensi ialah x = x/a, y = Re 1/2 ȳ/a, u = ū/u, v = Re 1/2 v/u, u e = ū e /U, T = Re 1/2 T T q w a/k, λ = Gr Re 5/2, dengan Re = U a/υ ialah nombor Reynolds, Gr = gβq w a 4 /kυ 2 ialah nombor Grasho dan λ ialah parameter olakan campuran. Perlu dinyatakan di sini bahawa λ > 0 mewakili aliran membantu dan λ < 0 mewakili aliran menentang. Nota: ψ ialah ungsi strim yang ditakrikan sebagai u = ψ and v = ψ x. a Dengan menggunakan pemboleh-pemboleh ubah tak serupa berbentuk berikut ψ = x x,y, T = θx,y, tunjukkan bahawa persamaan-persamaan lapisan sempadan 6 8 boleh diturunkan kepada sistem persamaan pembezaan berikut sinxcosx + λ sinx x x θ 2 = x x 2 x 2, /-
10 θ Pr 2 + θ = x θ x θ, 11 x dengan syarat-syarat sempadan 9 menjadi = = 0, θ = 1 pada y = 0, 12 sinx, g 0 apabila y. x 2 b Jadual 2 memaparkan nilai-nilai tegasan ricih 2 dan suhu permukaan y=0 θ y=0 bagi sistem persamaan 10 dan 11 tertakluk kepada syarat-syarat sempadan 12 bila x = 0 dan Pr = 7 dengan beberapa nilai λ. Tasirkan keputusan yang diperoleh dalam Jadual 2. Jadual 2: Nilai-nilai tegasan ricih dan suhu permukaan dengan beberapa nilai λ λ 2 2 y=0 θ y= [35 markah] - ooo O ooo -
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